Solve each of the following differential equations :?

Question

dy/dx+1/x y= (1-x)y^2

x^2 d^2x/dy^2 =2xdx/dy=0

Answer 1

(1)dy/dx+y/x=(1-x)y^2

this equation is of the form:
dy/dx+p(x)y=q(x)y^k

the substitution u=y^(-k+1)
transforms into;
du/dx+(-k+1)p(x)u
=(-k+1)q(x)...........(1)
here,k=2,p(x)=1/x,q(x)=(1-x)
let u=y^(-1)
substitute all this into (1)
du/dx+(-1)(1/x)u=(-1)(1-x)
du/dx-(1/x)u=(x-1)......(2)
this can now be solved by the
integrating factor method
for equations of the form;
du/dx+g(x)u=h(x)
g(x)= -1/x,h(x)=(x-1)
(a)the integrating factor
p=e^(int{-1/x}dx)
=e^(-ln(x))= 1/x
(b)d/dx(1/x)u=1/x(x-1)
(c)integrate
(1/x)u=int{(x-1)/x}dx
=int{1-1/x}dx
=x-lnx+C
therefore,u= x^2-xlnx+Cx
where C
is a constant
but, y=1/u,
hence,
y=1/(x^2-xlnx+Cx)

Answer 2

First ODE:

dy/dx + y/x = (1-x)y^2

Multiply both sides by 'x';

x(dy/dx) + y = x(1-x)y^2

The LHS can be written as:

d(xy)/dx

Thus, our eqn becomes:

d(xy)/dx = x(1-x)y^2

Assume the bracketed LHS term 'xy' equal to 'z';

dz/dx = (1-x)((xy)^2)/x

dz/dx = (1-x)(z^2)/x

(1/z^2)dz/dx = (1-x)/x

dz/z^2 = {(1-x)/x}dx = (dx/x) - dx

Integrating both sides,

-1/z = ln(x) - x + C

1/z = x - ln(x) + C

Back substituting for z, we get:

1/(xy) = x - ln(x) + C

1/y = x^2 - x*ln(x) + Cx

This is the solution to your ODE

Answer 3

4w² +12w - 40 4 = 0 w² + 3w -11 = 0...<---- divide by 4 to get first coefficient = a million w² + 3w + 9/4 - 9/4 - 40 4/4 = 0 ..<---- upload and subtract a million/2 of sq. of center coefficient (w + 3/2)² - fifty 3/4 = 0...<--- write as binomial squared and combine words (w + 3/2)² = fifty 3/4 ... <--- upload fifty 3/4 to the two facets w + 3/2 = ±(?fifty 3)/2.. <--- take sq. root w = (-3±?fifty 3) / 2.. <--- subtract 3/2 = -10.28 or 4.28 This one follows the comparable steps 3r² +18r = 21 r² + 6r = 7 r² + 6r + 9 = 7 + 9 (r + 3)² = sixteen r + 3 = ±4 r = -3±4 r = -7 or a million

Answer 4

1)
Bernoulli equation:
dy/dx+y/x = (1-x)y²
y'+y/x = (1-x)y² ........(*1/y²)
(1/y²)y'+1/(xy) = (1-x) ........... (*-1)
-(1/y²)y'-1/(xy) = (x-1)

=>
z = y^(1-2) = y^-1 = 1/y
z' = (-1/y²)y'

=>
-(1/y²)y'-1/(xy) = (x-1)
z' - z/x = (x-1)

Integrant factor:
I(x) = exp(∫(-1/x)dx) = exp(-1*ln x) = exp(ln x^-1) = x^-1 = 1/x

=>
z' - z/x = x-1 ....... (*1/x)
(1/x)*z' - (z/x)*(1/x) = (x-1)*(1/x)
z'/x - z/x² = (x-1)/x
(z/x)' = (x-1)/x
z/x = ∫[(x-1)/x]dx =∫1 dx -∫(1/x)dx = x - ln x + c
z = x² - x*ln x +cx
=>
1/y = x² - x*ln x
y = 1 / (x² - x*ln x + cx)

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